Chapter 7: Introduction to Trigonometry

Algebra 2 · Updated February 2026 · 35 min read

Trigonometry — literally "triangle measurement" — is one of the oldest and most practical branches of mathematics. Ancient astronomers used it to map the stars, engineers use it to design bridges and buildings, and physicists use it to model waves and oscillations. In this chapter you will learn the foundations of trigonometry: how the ratios of sides in a right triangle give rise to the sine, cosine, and tangent functions; how the unit circle extends these functions to all angles; how radians provide an alternative and often more natural way to measure angles; and how to graph, transform, and invert trigonometric functions. By the end of this chapter you will be prepared for the more advanced identities, equations, and applications that appear in Precalculus and Calculus.

7.1 Right Triangle Trigonometry

Trigonometry begins with right triangles. Given a right triangle with an acute angle $\theta$, we label the three sides relative to that angle: the side directly across from $\theta$ is the opposite side, the side adjacent to $\theta$ (that is not the hypotenuse) is the adjacent side, and the longest side — opposite the right angle — is the hypotenuse.

Definition: The Three Primary Trig Ratios (SOH CAH TOA)

For an acute angle $\theta$ in a right triangle:

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$

The mnemonic SOH CAH TOA encodes these: Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, Tangent = Opposite / Adjacent.

These ratios depend only on the angle $\theta$, not on the size of the triangle. Any two right triangles that share the same acute angle are similar, so their side ratios are identical. This is the fundamental insight that makes trigonometry so powerful: knowing just one angle and one side of a right triangle is enough to determine all the other sides.

Finding Missing Sides

To find an unknown side of a right triangle, choose the trig ratio that relates the known side, the unknown side, and the given angle. Then solve the resulting equation.

Example 7.1.1 — Finding a Missing Side

Problem: In a right triangle, the angle $\theta = 35°$ and the hypotenuse is $20$. Find the length of the side opposite $\theta$.

Step 1 Identify the relationship. We know the hypotenuse and want the opposite side, so we use sine:

$$\sin 35° = \frac{\text{opposite}}{20}$$

Step 2 Solve for the opposite side:

$$\text{opposite} = 20 \sin 35° \approx 20 \times 0.5736 \approx 11.47$$

Finding Missing Angles

When you know two sides of a right triangle and need to find an acute angle, compute the appropriate trig ratio and then use the inverse function on your calculator. For instance, if $\sin\theta = 0.6$, then $\theta = \sin^{-1}(0.6) \approx 36.87°$.

Example 7.1.2 — Finding an Angle

Problem: A right triangle has legs of length $5$ and $12$. Find the measure of the angle opposite the side of length $5$.

Step 1 We know both legs. The side of length $5$ is opposite our target angle, and $12$ is adjacent. Use tangent:

$$\tan\theta = \frac{5}{12}$$

Step 2 Apply the inverse tangent:

$$\theta = \tan^{-1}\!\left(\frac{5}{12}\right) \approx 22.62°$$

Special Right Triangles

Two families of right triangles appear so frequently in mathematics that their side ratios should be memorized. They come from bisecting an equilateral triangle and from cutting a square along its diagonal.

The 45-45-90 Triangle

In a $45°$-$45°$-$90°$ triangle, the two legs are equal and the hypotenuse is $\sqrt{2}$ times as long as each leg. If each leg has length $a$, the sides are:

$$a : a : a\sqrt{2}$$

Therefore: $\sin 45° = \cos 45° = \dfrac{\sqrt{2}}{2}$ and $\tan 45° = 1$.

The 30-60-90 Triangle

In a $30°$-$60°$-$90°$ triangle, if the shorter leg (opposite $30°$) has length $a$, then the longer leg (opposite $60°$) has length $a\sqrt{3}$ and the hypotenuse has length $2a$:

$$a : a\sqrt{3} : 2a$$

Therefore: $\sin 30° = \dfrac{1}{2}$, $\cos 30° = \dfrac{\sqrt{3}}{2}$, $\tan 30° = \dfrac{\sqrt{3}}{3}$, $\sin 60° = \dfrac{\sqrt{3}}{2}$, $\cos 60° = \dfrac{1}{2}$, $\tan 60° = \sqrt{3}$.

Example 7.1.3 — Using Special Triangles

Problem: A ladder leans against a wall making a $60°$ angle with the ground. The foot of the ladder is $4$ meters from the base of the wall. How long is the ladder?

Step 1 The $60°$ angle is between the ground and the ladder. The ground distance ($4$ m) is the side adjacent to the $60°$ angle. The ladder is the hypotenuse.

Step 2 Use cosine:

$$\cos 60° = \frac{4}{\text{hypotenuse}} \implies \frac{1}{2} = \frac{4}{\text{hyp}} \implies \text{hyp} = 8 \text{ meters}$$

Study Tip: Memorize the side ratios of the 30-60-90 and 45-45-90 triangles. These ratios recur throughout trigonometry, standardized tests, and college-level math. Being able to recall them instantly will save you considerable time.

7.2 The Unit Circle

Right triangle trigonometry only covers acute angles ($0° < \theta < 90°$). To extend trigonometric functions to all angles — including obtuse angles, negative angles, and angles greater than $360°$ — we use the unit circle, which is the circle of radius $1$ centered at the origin: $x^2 + y^2 = 1$.

Definition: Trig Functions on the Unit Circle

Place an angle $\theta$ in standard position: vertex at the origin, initial side along the positive $x$-axis. The terminal side of $\theta$ intersects the unit circle at a point $P(x, y)$. We define:

$$\cos\theta = x \qquad \sin\theta = y \qquad \tan\theta = \frac{y}{x} \; (x \neq 0)$$

This agrees with SOH CAH TOA for acute angles (where $\text{hyp} = 1$ on the unit circle) and extends naturally to all angles.

From the equation $x^2 + y^2 = 1$ and the definitions above, we immediately obtain the most important identity in trigonometry:

Pythagorean Identity

$$\sin^2\theta + \cos^2\theta = 1$$

This holds for every angle $\theta$. It is the trigonometric version of the Pythagorean Theorem.

Reference Angles

A reference angle is the acute angle formed between the terminal side of an angle and the $x$-axis. It always lies between $0°$ and $90°$. To evaluate trig functions at any angle, find its reference angle, look up the trig value of that reference angle, and then apply the correct sign based on the quadrant.

Quadrant Signs: All Students Take Calculus

The signs of sine, cosine, and tangent depend on which quadrant the terminal side falls in. Since $\cos\theta = x$ and $\sin\theta = y$, the signs follow from whether $x$ and $y$ are positive or negative in each quadrant.

Quadrant $\sin\theta$ ($y$) $\cos\theta$ ($x$) $\tan\theta$ ($y/x$) Mnemonic
I $+$ $+$ $+$ All positive
II $+$ $-$ $-$ Sine positive
III $-$ $-$ $+$ Tangent positive
IV $-$ $+$ $-$ Cosine positive

The mnemonic "All Students Take Calculus" reads counterclockwise from Quadrant I: All, Sine, Tangent, Cosine. Each word tells you which trig function(s) are positive in that quadrant.

Key Coordinates on the Unit Circle

Combining the special triangle values with the reference angle technique gives us exact coordinates for every multiple of $30°$ and $45°$ around the unit circle. Here are the values in the first quadrant; all other quadrants follow by adjusting signs.

Angle $0°$ $30°$ $45°$ $60°$ $90°$
$\cos\theta$ $1$ $\dfrac{\sqrt{3}}{2}$ $\dfrac{\sqrt{2}}{2}$ $\dfrac{1}{2}$ $0$
$\sin\theta$ $0$ $\dfrac{1}{2}$ $\dfrac{\sqrt{2}}{2}$ $\dfrac{\sqrt{3}}{2}$ $1$

Example 7.2.1 — Evaluating Trig Functions Using the Unit Circle

Problem: Find $\sin 225°$ and $\cos 225°$.

Step 1 $225°$ is in Quadrant III. Reference angle: $225° - 180° = 45°$.

Step 2 In Quadrant III, both sine and cosine are negative.

Step 3 Apply the reference angle values:

$$\sin 225° = -\sin 45° = -\frac{\sqrt{2}}{2} \qquad \cos 225° = -\cos 45° = -\frac{\sqrt{2}}{2}$$

Example 7.2.2 — Finding Trig Values from One Known Value

Problem: If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II, find $\cos\theta$ and $\tan\theta$.

Step 1 Use the Pythagorean identity:

$$\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$

Step 2 In Quadrant II, cosine is negative: $\cos\theta = -\dfrac{4}{5}$.

Step 3 Compute tangent:

$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$$

Interactive: The Unit Circle

Drag the slider below to move a point around the unit circle. Watch how the coordinates $(\cos\theta, \sin\theta)$ change as the angle $\theta$ varies.

The point traces the unit circle as the angle slider changes. The dashed lines show the sine and cosine projections.

7.3 Radian Measure

Degrees are intuitive but somewhat arbitrary — why divide a circle into $360$ parts? Radians offer a more natural and mathematically elegant way to measure angles. In calculus and higher mathematics, radians are the standard unit because they simplify many formulas.

Definition: Radian

One radian is the measure of a central angle that intercepts an arc whose length equals the radius of the circle. A full circle has circumference $2\pi r$, so a full rotation is $2\pi$ radians.

Converting Between Degrees and Radians

Since $360° = 2\pi$ radians, we have $180° = \pi$ radians. This gives the conversion factors:

Degree-Radian Conversion

$$\text{Degrees to radians: } \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}$$

$$\text{Radians to degrees: } \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi}$$

Example 7.3.1 — Converting Between Units

Problem: (a) Convert $150°$ to radians. (b) Convert $\dfrac{7\pi}{6}$ radians to degrees.

(a) $150° \times \dfrac{\pi}{180} = \dfrac{150\pi}{180} = \dfrac{5\pi}{6}$ radians.

(b) $\dfrac{7\pi}{6} \times \dfrac{180}{\pi} = \dfrac{7 \times 180}{6} = \dfrac{1260}{6} = 210°$.

Here are the most commonly encountered degree-radian equivalences:

Degrees $0°$ $30°$ $45°$ $60°$ $90°$ $120°$ $180°$ $270°$ $360°$
Radians $0$ $\dfrac{\pi}{6}$ $\dfrac{\pi}{4}$ $\dfrac{\pi}{3}$ $\dfrac{\pi}{2}$ $\dfrac{2\pi}{3}$ $\pi$ $\dfrac{3\pi}{2}$ $2\pi$

Arc Length

The relationship between an arc, its central angle, and the radius is beautifully simple when the angle is measured in radians.

Arc Length Formula

If a central angle $\theta$ (in radians) subtends an arc of length $s$ on a circle of radius $r$, then:

$$s = r\theta$$

This formula is the reason radians are preferred in science and engineering. If we used degrees, the formula would need an extra factor of $\frac{\pi}{180}$, making it less clean.

Example 7.3.2 — Arc Length

Problem: A circle has radius $10$ cm. Find the length of the arc intercepted by a central angle of $\dfrac{3\pi}{4}$ radians.

$$s = r\theta = 10 \times \frac{3\pi}{4} = \frac{30\pi}{4} = \frac{15\pi}{2} \approx 23.56 \text{ cm}$$

Sector Area

A sector is the "pie slice" region bounded by two radii and an arc. Its area is a fraction of the full circle's area, proportional to the central angle.

Sector Area Formula

The area of a sector with central angle $\theta$ (in radians) and radius $r$ is:

$$A = \frac{1}{2}r^2\theta$$

Example 7.3.3 — Sector Area

Problem: A sprinkler waters a circular sector with radius $8$ meters and central angle $120°$. Find the area of the watered region.

Step 1 Convert to radians: $120° \times \frac{\pi}{180} = \frac{2\pi}{3}$.

Step 2 Apply the formula:

$$A = \frac{1}{2}(8)^2 \left(\frac{2\pi}{3}\right) = \frac{1}{2}(64)\left(\frac{2\pi}{3}\right) = \frac{64\pi}{3} \approx 67.02 \text{ m}^2$$

Study Tip: Always check that your angle is in radians before using $s = r\theta$ or $A = \frac{1}{2}r^2\theta$. If the angle is given in degrees, convert it first. Using degrees directly in these formulas is the single most common error in this section.

7.4 Graphing Sine and Cosine

The graphs of $y = \sin x$ and $y = \cos x$ are smooth, continuous waves that repeat every $2\pi$ units. These wave patterns model countless real-world phenomena: sound waves, alternating current, tides, pendulum motion, and seasonal temperature cycles. Learning to read and transform these graphs is one of the most important skills in Algebra 2.

The Basic Graphs

The graph of $y = \sin x$ starts at the origin, rises to a peak of $1$ at $x = \frac{\pi}{2}$, returns to $0$ at $x = \pi$, drops to a trough of $-1$ at $x = \frac{3\pi}{2}$, and returns to $0$ at $x = 2\pi$ to complete one full cycle. The graph of $y = \cos x$ has the same shape but starts at its peak: it begins at $1$, drops to $0$ at $\frac{\pi}{2}$, reaches $-1$ at $\pi$, returns to $0$ at $\frac{3\pi}{2}$, and returns to $1$ at $2\pi$.

Key properties shared by both basic graphs:

The General Sinusoidal Form

The general form of a transformed sinusoidal function is:

$$y = A\sin\bigl(B(x - C)\bigr) + D$$

(The same form applies with cosine.) Each of the four parameters $A$, $B$, $C$, $D$ controls a specific transformation:

Parameter Name Effect Formula
$A$ Amplitude Vertical stretch or compression. If $A < 0$, the graph reflects over the midline. Amplitude $= |A|$
$B$ Frequency factor Horizontal stretch or compression, changing the period. Period $= \dfrac{2\pi}{|B|}$
$C$ Phase shift Horizontal translation. Positive $C$ shifts right; negative shifts left. Phase shift $= C$
$D$ Vertical shift Moves the midline up ($D > 0$) or down ($D < 0$). Midline: $y = D$

Common mistake: The argument must be factored as $B(x - C)$, not left as $Bx - C$. For instance, $y = \sin(2x - \pi)$ must be rewritten as $y = \sin\bigl(2(x - \frac{\pi}{2})\bigr)$ before reading off the phase shift. The phase shift is $\frac{\pi}{2}$ to the right, not $\pi$.

Example 7.4.1 — Identifying Parameters

Problem: For $y = -3\cos(4x + \pi) + 2$, identify the amplitude, period, phase shift, and midline.

Step 1 Factor the argument: $4x + \pi = 4\bigl(x + \frac{\pi}{4}\bigr) = 4\bigl(x - (-\frac{\pi}{4})\bigr)$.

So $A = -3$, $B = 4$, $C = -\frac{\pi}{4}$, $D = 2$.

Step 2 Compute:

Example 7.4.2 — Writing an Equation from a Description

Problem: A sinusoidal function has amplitude $5$, period $6\pi$, phase shift $\frac{\pi}{3}$ to the right, and midline $y = -1$. Write a sine equation.

Step 1 $A = 5$, $C = \frac{\pi}{3}$, $D = -1$.

Step 2 Find $B$: Period $= \frac{2\pi}{|B|} = 6\pi \implies |B| = \frac{2\pi}{6\pi} = \frac{1}{3}$.

Step 3 Assemble:

$$y = 5\sin\!\left(\frac{1}{3}\left(x - \frac{\pi}{3}\right)\right) - 1$$

Interactive: Sine/Cosine Transformer

Use the sliders below to adjust $A$, $B$, $C$, and $D$ and watch how the graph of $y = A\sin(B(x - C)) + D$ (blue) transforms compared to the base graph $y = \sin x$ (dashed gray). The dashed green line shows the midline $y = D$.

Drag the sliders to explore amplitude, period, phase shift, and vertical shift in real time.

7.5 Graphing Tangent and Other Trig Functions

The tangent, cotangent, secant, and cosecant functions have graphs that look very different from the smooth waves of sine and cosine. Their defining feature is the presence of vertical asymptotes where the function is undefined.

The Tangent Function

Since $\tan x = \frac{\sin x}{\cos x}$, the function is undefined wherever $\cos x = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. At these values the graph has vertical asymptotes.

Properties of $y = \tan x$

Domain: All real numbers except $x = \frac{\pi}{2} + n\pi$

Range: $(-\infty, \infty)$

Period: $\pi$ (half the period of sine and cosine)

No amplitude (the function is unbounded)

The graph passes through the origin, is increasing on each branch, and has an inflection point at each $x$-intercept.

For a transformed tangent $y = A\tan(Bx)$, the period becomes $\frac{\pi}{|B|}$ and the vertical stretch factor is $|A|$. Vertical asymptotes shift according to the transformation.

Example 7.5.1 — Graphing a Transformed Tangent

Problem: Find the period and two consecutive asymptotes of $y = \tan(3x)$, and sketch one period.

Step 1 Period $= \frac{\pi}{|B|} = \frac{\pi}{3}$.

Step 2 Asymptotes occur where $3x = \frac{\pi}{2} + n\pi$, i.e., $x = \frac{\pi}{6} + \frac{n\pi}{3}$.

Two consecutive asymptotes: $x = -\frac{\pi}{6}$ (when $n = -1$) and $x = \frac{\pi}{6}$ (when $n = 0$).

Step 3 Key points within this period:

The graph rises from $-\infty$ to $+\infty$ between the two asymptotes, passing through these three points.

Cotangent, Secant, and Cosecant

The remaining three trig functions are reciprocals of the primary three:

Function Period Asymptotes Range
$\tan x$ $\pi$ $x = \frac{\pi}{2} + n\pi$ $(-\infty, \infty)$
$\cot x$ $\pi$ $x = n\pi$ $(-\infty, \infty)$
$\sec x$ $2\pi$ $x = \frac{\pi}{2} + n\pi$ $(-\infty, -1] \cup [1, \infty)$
$\csc x$ $2\pi$ $x = n\pi$ $(-\infty, -1] \cup [1, \infty)$

Graphing Tip: To graph $\csc x$ or $\sec x$, first sketch the corresponding reciprocal function ($\sin x$ or $\cos x$) as a dashed guide curve. Wherever the guide curve crosses the $x$-axis, draw a vertical asymptote. The branches of the reciprocal function open away from the $x$-axis, touching the guide curve at its peaks and troughs.

Example 7.5.2 — Period and Asymptotes of Cosecant

Problem: Find the period and vertical asymptotes of $y = \csc(2x)$.

Step 1 The period of $\csc(Bx)$ is $\frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$.

Step 2 Vertical asymptotes occur where $\sin(2x) = 0$, i.e., $2x = n\pi$, so $x = \frac{n\pi}{2}$.

The asymptotes are at $x = 0, \pm\frac{\pi}{2}, \pm\pi, \pm\frac{3\pi}{2}, \ldots$

7.6 Inverse Trigonometric Functions

Trigonometric functions are periodic, so they fail the Horizontal Line Test on their full domains and are not one-to-one. To define inverse functions, we must restrict each trig function to an interval where it is one-to-one. The resulting inverses "undo" the trig functions and are essential for solving trigonometric equations.

The Three Principal Inverse Trig Functions

Arcsine: $y = \arcsin x = \sin^{-1} x$ means $\sin y = x$ with $-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$.

Domain: $[-1, 1]$.   Range: $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Arccosine: $y = \arccos x = \cos^{-1} x$ means $\cos y = x$ with $0 \leq y \leq \pi$.

Domain: $[-1, 1]$.   Range: $[0, \pi]$.

Arctangent: $y = \arctan x = \tan^{-1} x$ means $\tan y = x$ with $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$.

Domain: $(-\infty, \infty)$.   Range: $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Notation caution: The notation $\sin^{-1}x$ means the inverse sine function (arcsine), not $\frac{1}{\sin x}$. To write the reciprocal, use $(\sin x)^{-1}$ or $\csc x$. This notational trap catches many students.

Evaluating Inverse Trig Functions

To evaluate an expression like $\arcsin\!\left(\frac{1}{2}\right)$, ask: "What angle $\theta$ in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ satisfies $\sin\theta = \frac{1}{2}$?" Since $\sin\frac{\pi}{6} = \frac{1}{2}$ and $\frac{\pi}{6}$ is in the required range, the answer is $\frac{\pi}{6}$ (or equivalently, $30°$).

Example 7.6.1 — Evaluating Inverse Trig Expressions

Problem: Find the exact values of: (a) $\arccos\!\left(-\dfrac{\sqrt{2}}{2}\right)$, (b) $\arctan(1)$, (c) $\arcsin(-1)$.

(a) We need $\theta \in [0, \pi]$ with $\cos\theta = -\frac{\sqrt{2}}{2}$. The reference angle is $\frac{\pi}{4}$ (since $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$). Negative cosine means Quadrant II: $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

(b) We need $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan\theta = 1$. Since $\tan\frac{\pi}{4} = 1$, the answer is $\frac{\pi}{4}$.

(c) We need $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin\theta = -1$. Since $\sin\!\left(-\frac{\pi}{2}\right) = -1$, the answer is $-\frac{\pi}{2}$.

Compositions of Trig and Inverse Trig Functions

A common type of problem asks you to evaluate a trig function of an inverse trig function, such as $\cos(\arcsin x)$. The strategy is to let $\theta$ equal the inner inverse trig expression, draw a right triangle, label the sides, and then read off the desired trig value.

Example 7.6.2 — Evaluating a Composition

Problem: Find $\cos\!\left(\arctan\dfrac{5}{12}\right)$.

Step 1 Let $\theta = \arctan\frac{5}{12}$. Then $\tan\theta = \frac{5}{12}$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Since the value is positive, $\theta$ is in Quadrant I.

Step 2 In the right triangle for $\theta$: opposite $= 5$, adjacent $= 12$. By the Pythagorean Theorem, hypotenuse $= \sqrt{25 + 144} = \sqrt{169} = 13$.

Step 3 Read off cosine:

$$\cos\!\left(\arctan\frac{5}{12}\right) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$$

Example 7.6.3 — Another Composition

Problem: Find $\sin\!\left(\arccos\!\left(-\dfrac{3}{5}\right)\right)$.

Step 1 Let $\theta = \arccos\!\left(-\frac{3}{5}\right)$. Then $\cos\theta = -\frac{3}{5}$ with $\theta \in [0, \pi]$. Since cosine is negative, $\theta$ is in Quadrant II, where sine is positive.

Step 2 Use the Pythagorean identity:

$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$

We take the positive root because $\theta$ is in Quadrant II.

Study Tip: When evaluating compositions, the "triangle method" always works. Let $\theta$ equal the inner inverse trig expression, draw a right triangle with the appropriate sides, find the missing side using the Pythagorean Theorem, and then read off whatever trig ratio the outer function asks for. Pay attention to the quadrant to get the sign right.

7.7 Practice Problems

Test your understanding of Chapter 7 with these problems. Click "Show Solution" to reveal a full worked solution for each one.

Problem 1

In a right triangle, one acute angle is $40°$ and the adjacent side is $15$. Find the opposite side.

Show Solution

We know the adjacent side and want the opposite side, so we use tangent:

$$\tan 40° = \frac{\text{opposite}}{15} \implies \text{opposite} = 15\tan 40° \approx 15 \times 0.8391 \approx 12.59$$

Problem 2

A 45-45-90 triangle has a hypotenuse of length $10\sqrt{2}$. Find the length of each leg.

Show Solution

In a 45-45-90 triangle, the sides are in the ratio $a : a : a\sqrt{2}$. So the hypotenuse $= a\sqrt{2}$.

$$a\sqrt{2} = 10\sqrt{2} \implies a = 10$$

Each leg has length $10$.

Problem 3

Find the exact values of $\sin 315°$ and $\cos 315°$ using the unit circle.

Show Solution

$315°$ is in Quadrant IV. Reference angle: $360° - 315° = 45°$.

In Quadrant IV, cosine is positive and sine is negative:

$$\sin 315° = -\sin 45° = -\frac{\sqrt{2}}{2}$$

$$\cos 315° = +\cos 45° = \frac{\sqrt{2}}{2}$$

Problem 4

If $\cos\theta = -\dfrac{5}{13}$ and $\theta$ is in Quadrant III, find $\sin\theta$ and $\tan\theta$.

Show Solution

Use $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$.

In Quadrant III, sine is negative: $\sin\theta = -\frac{12}{13}$.

$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-12/13}{-5/13} = \frac{12}{5}$$

(Tangent is positive in Quadrant III, as expected.)

Problem 5

Convert $\dfrac{5\pi}{12}$ radians to degrees, and convert $225°$ to radians.

Show Solution

Radians to degrees: $\dfrac{5\pi}{12} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{12} = \dfrac{900}{12} = 75°$.

Degrees to radians: $225° \times \dfrac{\pi}{180} = \dfrac{225\pi}{180} = \dfrac{5\pi}{4}$.

Problem 6

A circle has radius $6$ cm. Find the arc length and sector area for a central angle of $\dfrac{2\pi}{3}$ radians.

Show Solution

Arc length: $s = r\theta = 6 \times \frac{2\pi}{3} = 4\pi \approx 12.57$ cm.

Sector area: $A = \frac{1}{2}r^2\theta = \frac{1}{2}(36)\left(\frac{2\pi}{3}\right) = 12\pi \approx 37.70$ cm$^2$.

Problem 7

Find the amplitude, period, phase shift, and midline of $y = 4\sin\!\left(2x - \dfrac{\pi}{3}\right) - 1$.

Show Solution

Rewrite: $y = 4\sin\!\left(2\!\left(x - \frac{\pi}{6}\right)\right) - 1$.

Parameters: $A = 4$, $B = 2$, $C = \frac{\pi}{6}$, $D = -1$.

  • Amplitude: $|A| = 4$
  • Period: $\frac{2\pi}{2} = \pi$
  • Phase shift: $\frac{\pi}{6}$ to the right
  • Midline: $y = -1$

Problem 8

Write a cosine equation with amplitude $2$, period $8\pi$, phase shift $\pi$ to the right, and midline $y = 3$.

Show Solution

$A = 2$, $C = \pi$, $D = 3$.

Period $= \frac{2\pi}{|B|} = 8\pi \implies |B| = \frac{2\pi}{8\pi} = \frac{1}{4}$.

$$y = 2\cos\!\left(\frac{1}{4}(x - \pi)\right) + 3$$

Problem 9

Find the period and two consecutive vertical asymptotes of $y = \tan\!\left(\dfrac{\pi x}{2}\right)$.

Show Solution

Period $= \frac{\pi}{|B|} = \frac{\pi}{\pi/2} = 2$.

Asymptotes occur where $\frac{\pi x}{2} = \frac{\pi}{2} + n\pi$, i.e., $x = 1 + 2n$.

Two consecutive asymptotes: $x = -1$ (when $n = -1$) and $x = 1$ (when $n = 0$).

Problem 10

Evaluate: (a) $\arcsin\!\left(\dfrac{\sqrt{3}}{2}\right)$,   (b) $\arccos(0)$,   (c) $\arctan(-\sqrt{3})$.

Show Solution

(a) We need $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin\theta = \frac{\sqrt{3}}{2}$. Answer: $\theta = \frac{\pi}{3}$ (i.e., $60°$).

(b) We need $\theta \in [0, \pi]$ with $\cos\theta = 0$. Answer: $\theta = \frac{\pi}{2}$ (i.e., $90°$).

(c) We need $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan\theta = -\sqrt{3}$. The reference angle is $\frac{\pi}{3}$ (since $\tan\frac{\pi}{3} = \sqrt{3}$). The negative sign places us in the fourth-quadrant portion of the range: $\theta = -\frac{\pi}{3}$.

Problem 11

Find the exact value of $\sin\!\left(\arccos\dfrac{4}{5}\right)$.

Show Solution

Let $\theta = \arccos\frac{4}{5}$. Then $\cos\theta = \frac{4}{5}$ with $\theta \in [0, \pi]$. Since $\cos\theta > 0$, $\theta$ is in Quadrant I where sine is positive.

$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

Problem 12

A Ferris wheel has a diameter of $50$ meters with its center $30$ meters above the ground. It completes one revolution every $4$ minutes. Write a sinusoidal equation for the height $h(t)$ of a rider (in meters) as a function of time $t$ (in minutes), assuming the rider starts at the lowest point.

Show Solution

The radius (amplitude) is $A = 25$. The midline (center height) is $D = 30$. The period is $4$ minutes, so $B = \frac{2\pi}{4} = \frac{\pi}{2}$.

Starting at the bottom means starting at the minimum. A negative cosine starts at its minimum:

$$h(t) = -25\cos\!\left(\frac{\pi}{2}\,t\right) + 30$$

Check: At $t = 0$: $h(0) = -25(1) + 30 = 5$ m (bottom, since center is 30 m and radius is 25 m). At $t = 2$: $h(2) = -25(-1) + 30 = 55$ m (top). Correct.

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